Intersection of Random Subsets Consider $n$ independently drawn $q$-subsets $e_1,...,e_n$ from a finite set $P$. What is the probability that the intersection of the $n$ subsets is non-empty, in terms of $n$, $q$, and $|P|$?
$$\mathbb{P}\left[\bigcap_{i=1}^n e_i \neq \emptyset \right] = \,\,?$$
EDIT: I was asked to give my thoughts on the problem. It comes up from a bit of research I am doing on edge colorings of graphs with no bichromatic cycles. I am not sure how to attack this type of problem. I should have stated: I am not necessarily looking for an answer. Any insight or suggestions on how to proceed would be welcome.
 A: Credit goes to Studentmath and Ted Shifrin for discussing this with me in the MSE chat.

One of the most viable approaches (even though not exactly elegant) is to apply inclusion-exclusion on the size of the intersection $\displaystyle\bigcap_{i=1}^n e_i$.
So we determine the probability that the intersection contains a certain set of size $i$, say. To do this, we choose $i$ elements, and then for each $q$-subset, $q-i$ elements to go with it. This yields: $$\binom p i \binom{p-i}{q-i}^n \binom{p}{q}^{-n}$$
Now, of course, we have to do the familiar correction for double-counting, yielding the following inclusion-exclusion summation: $$\sum_{i=1}^q (-1)^{i+1} \binom p i \binom{p-i}{q-i}^n \binom{p}{q}^{-n}$$

Update: When there is a desire to calculate multiple values, or to know the exact distribution over the different intersection sizes, the following recursive approach may be useful:
Let $N(k, i)$ denote the number of ways $k$ $q$-subsets can have an intersection with $i$ elements. Then we can derive $N(k, i)$ from the $N(k-1, *)$ as follows:
\begin{align*}
N(k,i) &= \frac 1n \sum_{j=i}^q N(k-1,j) \binom j i \binom{p-j}{q-i} \\
N(1,i) &= \begin{cases}
0 & :i \ne q \\
\binom p q & :i = q
\end{cases}
\end{align*}
where the $\frac 1n$ corrects for the otherwise ordered sequence of adding the $q$-subsets to our consideration.
A: I don't believe that Lord_Farin's first solution is complete. I've been exploring this problem for the past few days, and that solution as is doesn't provide correct answers according to the brute force computations I calculated for comparison. However, if you start the summation at $i=0$ instead of $i=1$ and use $(-1)^i$ instead of $(-1)^{i+1}$ it will give the correct probability of no overlap.
I do have a recursive solution that is similar to the one they Lord_Farin gave, though the only thing keeping me from getting rid of the recursion was that I didn't have a closed-form expression for the probability of no overlap, which I now have thanks to Lord_Farin: 
$$\mathbb{P}(i|p,q,n) = {p \choose i} {p-i \choose q-i}^{n} {p \choose q}^{-n} \sum_{j=0}^{q-i} (-1)^{j} {p-i \choose j} {p-i-j \choose q-i-j}^{n} {p-i \choose q-i}^{-n} $$
That should be a closed form expression for the probability of getting any given overlap of size $i$ for any $0\le i \le q$.
In words, we have ${p \choose i}$ ways of choosing ${i}$ elements from ${p}$ to form an overlap of size ${i}$, ${p-i \choose q-i}^n$ ways of choosing the remaining ${q-i}$ elements to fill out each set from the ${p-i}$ elements not in the overlap, and ${p \choose q}^n$ possible collections of sets in our sample space. However, some of the choices for filling out each set will add to the overlap, which we don't want. We only want to choose the $n$ smaller $(q-i)-$subsets such that they have no overlap among themselves. Using the inclusion-exclusion principle, we can calculate what fraction of these will have an empty overlap. The final summation term does this.
A: Another way to approach this problem is to invoke De Morgan's law and instead look at the cardinality of the union of the complements $\vert \displaystyle\bigcup_{k=1}^n \bar{e_k} \vert$. The $\bar{e_k}$ are now random subsets with cardinality $p-q$
We can then directly apply this convenient result: https://www.matem.unam.mx/~barot/articles/indirstat.pdf (DOI 10.1007/PL00000552 )
This gives:
$$\mathbb{P}(\vert \displaystyle\bigcap_{k=1}^n e_k \vert = i) = f_{p, q, n}(i) = \binom{p}{i}\binom{p}{q}^{-n}\sum_{j=0}^{q-i}(-1)^j\binom{p-i}{j}\binom{p-i-j}{p-q}^n$$
This is the same as Aaron Mishtal's answer (after simplification). I agree that Lord_Farin's answer is incorrect.
The answer to the original question is then:
$$\mathbb{P}\left[\bigcap_{k=1}^n e_k \neq \emptyset \right] = 1 - f_{p, q, n}(0) = 1 - \binom{p}{q}^{-n}\sum_{j=0}^{q-i}(-1)^j\binom{p}{j}\binom{p-j}{p-q}^n$$
